Mixing
Partial Differential Equations Question: State if the following PDEs are linear homogeneous, linear...
$begingroup$
State if the following PDEs are linear homogeneous, linear nonhomogeneous, or nonlinear:
$u_{t}+u_{x}=sin(x)u$
$u_{tt}-u_{xx}=e^{t}u_{t}$
$u_{tt}-u_{xx}=x^{2}$
$u_{xx}+u_{yy}=u_{x}u_{y}$
As for the first equation, I think it's linear homogeneous, and the second one is linear non-homogeneous...but I'm not sure about the last two! Would appreciate any help! Thank you in advance!
pde homogeneous-equation
asked Jan 20 at 21:33
JennyJenny
624
$endgroup$
add a comment |
$begingroup$
State if the following PDEs are linear homogeneous, linear nonhomogeneous, or nonlinear:
$u_{t}+u_{x}=sin(x)u$
$u_{tt}-u_{xx}=e^{t}u_{t}$
$u_{tt}-u_{xx}=x^{2}$
$u_{xx}+u_{yy}=u_{x}u_{y}$
As for the first equation, I think it's linear homogeneous, and the second one is linear non-homogeneous...but I'm not sure about the last two! Would appreciate any help! Thank you in advance!
pde homogeneous-equation
asked Jan 20 at 21:33
JennyJenny
624
$endgroup$
add a comment |
$begingroup$
State if the following PDEs are linear homogeneous, linear nonhomogeneous, or nonlinear:
$u_{t}+u_{x}=sin(x)u$
$u_{tt}-u_{xx}=e^{t}u_{t}$
$u_{tt}-u_{xx}=x^{2}$
$u_{xx}+u_{yy}=u_{x}u_{y}$
As for the first equation, I think it's linear homogeneous, and the second one is linear non-homogeneous...but I'm not sure about the last two! Would appreciate any help! Thank you in advance!
pde homogeneous-equation
asked Jan 20 at 21:33
JennyJenny
624
$endgroup$
State if the following PDEs are linear homogeneous, linear nonhomogeneous, or nonlinear:
$u_{t}+u_{x}=sin(x)u$
$u_{tt}-u_{xx}=e^{t}u_{t}$
$u_{tt}-u_{xx}=x^{2}$
$u_{xx}+u_{yy}=u_{x}u_{y}$
As for the first equation, I think it's linear homogeneous, and the second one is linear non-homogeneous...but I'm not sure about the last two! Would appreciate any help! Thank you in advance!
pde homogeneous-equation
pde homogeneous-equation
asked Jan 20 at 21:33
JennyJenny
624
asked Jan 20 at 21:33
JennyJenny
624
asked Jan 20 at 21:33
JennyJenny
624
asked Jan 20 at 21:33
JennyJenny
624
asked Jan 20 at 21:33
JennyJenny
624
624
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Keep it simple.
First group all the terms involving $u$ to get an expression of the form
$$
mathcal{D}(u)=f(t,x,y)
$$
if $f(t,x,y)=0$ your equation is said homogeneous, otherwise it is said non-homogeneous
Then consider the homogeneous equation $mathcal{D}(u)=0$ (ignoring the eventual term $f$). By definition the equation is said linear if given two solutions $u, v$ and two scalars $alpha, beta$ then $alpha u+beta v$ is also a solution. In other term you must check that $mathcal{D}(alpha u+beta v)=alpha mathcal{D}(u) + beta mathcal{D}(v)$. If the equation is not linear it is said non-linear.
$mathcal{D}(u)=u_t+u_x-sin(x)u$ and $f=0$, thus linear homogeneous.
$mathcal{D}(u)=u_{tt}-u_{xx}-e^tu_t$ and $f=0$, thus linear homogeneous.
$mathcal{D}(u)=u_{tt}-u_{xx}$ and $f=x^2$, thus linear non-homogeneous.
$mathcal{D}(u)=u_{xx}-u_{yy}-u_xu_y$ and $f=0$ , thus non-linear homogeneous.
answered Jan 20 at 22:01
Picaud VincentPicaud Vincent
1,434310
$endgroup$
add a comment |
$begingroup$
A linear PDE is one that is of first degree in all of its field variables and partial derivatives. Hence, all of them are linear.
$u_{t}+u_{x}=sin(x)u$
LINEAR HOMOGENEOUS.
$u_{tt}-u_{xx}=e^{t}u_{t}$
LINEAR HOMOGENEOUS.
$u_{tt}-u_{xx}=x^{2}$
LINEAR NON-HOMOGENEOUS.
$u_{xx}+u_{yy}=u_{x}u_{y}$
LINEAR HOMOGENEOUS.
answered Jan 20 at 21:47
idriskameniidriskameni
753321
$endgroup$
add a comment |
$begingroup$
PDEs are linear if there is a linear operator $mathscr{D}$ that operates only on $u$, and they are homogeneous if all terms contained within the PDE are accounted for in $mathscr{D}u.$
$$
mathscr{D}u = 0 text{ is homogeneous} \
mathscr{D}u = f text{ is nonhomogeneous}.
$$
answered Jan 20 at 21:48
AEngineerAEngineer
1,5441317
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Keep it simple.
First group all the terms involving $u$ to get an expression of the form
$$
mathcal{D}(u)=f(t,x,y)
$$
if $f(t,x,y)=0$ your equation is said homogeneous, otherwise it is said non-homogeneous
Then consider the homogeneous equation $mathcal{D}(u)=0$ (ignoring the eventual term $f$). By definition the equation is said linear if given two solutions $u, v$ and two scalars $alpha, beta$ then $alpha u+beta v$ is also a solution. In other term you must check that $mathcal{D}(alpha u+beta v)=alpha mathcal{D}(u) + beta mathcal{D}(v)$. If the equation is not linear it is said non-linear.
$mathcal{D}(u)=u_t+u_x-sin(x)u$ and $f=0$, thus linear homogeneous.
$mathcal{D}(u)=u_{tt}-u_{xx}-e^tu_t$ and $f=0$, thus linear homogeneous.
$mathcal{D}(u)=u_{tt}-u_{xx}$ and $f=x^2$, thus linear non-homogeneous.
$mathcal{D}(u)=u_{xx}-u_{yy}-u_xu_y$ and $f=0$ , thus non-linear homogeneous.
answered Jan 20 at 22:01
Picaud VincentPicaud Vincent
1,434310
$endgroup$
add a comment |
$begingroup$
Keep it simple.
First group all the terms involving $u$ to get an expression of the form
$$
mathcal{D}(u)=f(t,x,y)
$$
if $f(t,x,y)=0$ your equation is said homogeneous, otherwise it is said non-homogeneous
Then consider the homogeneous equation $mathcal{D}(u)=0$ (ignoring the eventual term $f$). By definition the equation is said linear if given two solutions $u, v$ and two scalars $alpha, beta$ then $alpha u+beta v$ is also a solution. In other term you must check that $mathcal{D}(alpha u+beta v)=alpha mathcal{D}(u) + beta mathcal{D}(v)$. If the equation is not linear it is said non-linear.
$mathcal{D}(u)=u_t+u_x-sin(x)u$ and $f=0$, thus linear homogeneous.
$mathcal{D}(u)=u_{tt}-u_{xx}-e^tu_t$ and $f=0$, thus linear homogeneous.
$mathcal{D}(u)=u_{tt}-u_{xx}$ and $f=x^2$, thus linear non-homogeneous.
$mathcal{D}(u)=u_{xx}-u_{yy}-u_xu_y$ and $f=0$ , thus non-linear homogeneous.
answered Jan 20 at 22:01
Picaud VincentPicaud Vincent
1,434310
$endgroup$
add a comment |
$begingroup$
Keep it simple.
First group all the terms involving $u$ to get an expression of the form
$$
mathcal{D}(u)=f(t,x,y)
$$
if $f(t,x,y)=0$ your equation is said homogeneous, otherwise it is said non-homogeneous
Then consider the homogeneous equation $mathcal{D}(u)=0$ (ignoring the eventual term $f$). By definition the equation is said linear if given two solutions $u, v$ and two scalars $alpha, beta$ then $alpha u+beta v$ is also a solution. In other term you must check that $mathcal{D}(alpha u+beta v)=alpha mathcal{D}(u) + beta mathcal{D}(v)$. If the equation is not linear it is said non-linear.
$mathcal{D}(u)=u_t+u_x-sin(x)u$ and $f=0$, thus linear homogeneous.
$mathcal{D}(u)=u_{tt}-u_{xx}-e^tu_t$ and $f=0$, thus linear homogeneous.
$mathcal{D}(u)=u_{tt}-u_{xx}$ and $f=x^2$, thus linear non-homogeneous.
$mathcal{D}(u)=u_{xx}-u_{yy}-u_xu_y$ and $f=0$ , thus non-linear homogeneous.
answered Jan 20 at 22:01
Picaud VincentPicaud Vincent
1,434310
$endgroup$
Keep it simple.
First group all the terms involving $u$ to get an expression of the form
$$
mathcal{D}(u)=f(t,x,y)
$$
if $f(t,x,y)=0$ your equation is said homogeneous, otherwise it is said non-homogeneous
Then consider the homogeneous equation $mathcal{D}(u)=0$ (ignoring the eventual term $f$). By definition the equation is said linear if given two solutions $u, v$ and two scalars $alpha, beta$ then $alpha u+beta v$ is also a solution. In other term you must check that $mathcal{D}(alpha u+beta v)=alpha mathcal{D}(u) + beta mathcal{D}(v)$. If the equation is not linear it is said non-linear.
$mathcal{D}(u)=u_t+u_x-sin(x)u$ and $f=0$, thus linear homogeneous.
$mathcal{D}(u)=u_{tt}-u_{xx}-e^tu_t$ and $f=0$, thus linear homogeneous.
$mathcal{D}(u)=u_{tt}-u_{xx}$ and $f=x^2$, thus linear non-homogeneous.
$mathcal{D}(u)=u_{xx}-u_{yy}-u_xu_y$ and $f=0$ , thus non-linear homogeneous.
answered Jan 20 at 22:01
Picaud VincentPicaud Vincent
1,434310
edited Jan 20 at 22:15
answered Jan 20 at 22:01
Picaud VincentPicaud Vincent
1,434310
answered Jan 20 at 22:01
Picaud VincentPicaud Vincent
1,434310
answered Jan 20 at 22:01
Picaud VincentPicaud Vincent
1,434310
1,434310
add a comment |
add a comment |
$begingroup$
A linear PDE is one that is of first degree in all of its field variables and partial derivatives. Hence, all of them are linear.
$u_{t}+u_{x}=sin(x)u$
LINEAR HOMOGENEOUS.
$u_{tt}-u_{xx}=e^{t}u_{t}$
LINEAR HOMOGENEOUS.
$u_{tt}-u_{xx}=x^{2}$
LINEAR NON-HOMOGENEOUS.
$u_{xx}+u_{yy}=u_{x}u_{y}$
LINEAR HOMOGENEOUS.
answered Jan 20 at 21:47
idriskameniidriskameni
753321
$endgroup$
add a comment |
$begingroup$
A linear PDE is one that is of first degree in all of its field variables and partial derivatives. Hence, all of them are linear.
$u_{t}+u_{x}=sin(x)u$
LINEAR HOMOGENEOUS.
$u_{tt}-u_{xx}=e^{t}u_{t}$
LINEAR HOMOGENEOUS.
$u_{tt}-u_{xx}=x^{2}$
LINEAR NON-HOMOGENEOUS.
$u_{xx}+u_{yy}=u_{x}u_{y}$
LINEAR HOMOGENEOUS.
answered Jan 20 at 21:47
idriskameniidriskameni
753321
$endgroup$
add a comment |
$begingroup$
A linear PDE is one that is of first degree in all of its field variables and partial derivatives. Hence, all of them are linear.
$u_{t}+u_{x}=sin(x)u$
LINEAR HOMOGENEOUS.
$u_{tt}-u_{xx}=e^{t}u_{t}$
LINEAR HOMOGENEOUS.
$u_{tt}-u_{xx}=x^{2}$
LINEAR NON-HOMOGENEOUS.
$u_{xx}+u_{yy}=u_{x}u_{y}$
LINEAR HOMOGENEOUS.
answered Jan 20 at 21:47
idriskameniidriskameni
753321
$endgroup$
A linear PDE is one that is of first degree in all of its field variables and partial derivatives. Hence, all of them are linear.
$u_{t}+u_{x}=sin(x)u$
LINEAR HOMOGENEOUS.
$u_{tt}-u_{xx}=e^{t}u_{t}$
LINEAR HOMOGENEOUS.
$u_{tt}-u_{xx}=x^{2}$
LINEAR NON-HOMOGENEOUS.
$u_{xx}+u_{yy}=u_{x}u_{y}$
LINEAR HOMOGENEOUS.
answered Jan 20 at 21:47
idriskameniidriskameni
753321
answered Jan 20 at 21:47
idriskameniidriskameni
753321
answered Jan 20 at 21:47
idriskameniidriskameni
753321
answered Jan 20 at 21:47
idriskameniidriskameni
753321
753321
add a comment |
add a comment |
$begingroup$
PDEs are linear if there is a linear operator $mathscr{D}$ that operates only on $u$, and they are homogeneous if all terms contained within the PDE are accounted for in $mathscr{D}u.$
$$
mathscr{D}u = 0 text{ is homogeneous} \
mathscr{D}u = f text{ is nonhomogeneous}.
$$
answered Jan 20 at 21:48
AEngineerAEngineer
1,5441317
$endgroup$
add a comment |
$begingroup$
PDEs are linear if there is a linear operator $mathscr{D}$ that operates only on $u$, and they are homogeneous if all terms contained within the PDE are accounted for in $mathscr{D}u.$
$$
mathscr{D}u = 0 text{ is homogeneous} \
mathscr{D}u = f text{ is nonhomogeneous}.
$$
answered Jan 20 at 21:48
AEngineerAEngineer
1,5441317
$endgroup$
add a comment |
$begingroup$
PDEs are linear if there is a linear operator $mathscr{D}$ that operates only on $u$, and they are homogeneous if all terms contained within the PDE are accounted for in $mathscr{D}u.$
$$
mathscr{D}u = 0 text{ is homogeneous} \
mathscr{D}u = f text{ is nonhomogeneous}.
$$
answered Jan 20 at 21:48
AEngineerAEngineer
1,5441317
$endgroup$
PDEs are linear if there is a linear operator $mathscr{D}$ that operates only on $u$, and they are homogeneous if all terms contained within the PDE are accounted for in $mathscr{D}u.$
$$
mathscr{D}u = 0 text{ is homogeneous} \
mathscr{D}u = f text{ is nonhomogeneous}.
$$
answered Jan 20 at 21:48
AEngineerAEngineer
1,5441317
answered Jan 20 at 21:48
AEngineerAEngineer
1,5441317
answered Jan 20 at 21:48
AEngineerAEngineer
1,5441317
answered Jan 20 at 21:48
AEngineerAEngineer
1,5441317
1,5441317
add a comment |
add a comment |
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